est the local bank’s claim. The population variances are not equal. Use α = 0.05
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- A local bank claims that the waiting time for its customers to be served is the lowest in the area. A competitor’s bank checks the waiting times at both banks. A random sample of 15 customers had a mean waiting time of 5.3 minutes with a standard deviation of 1.1 minutes. Sixteen randomly selected customers from the competitor’s bank had a mean of 5.6 minutes with a standard deviation of 1.0 minutes. Test the local bank’s claim. The population variances are not equal. Use α = 0.05
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- According to a news program, Americans take an average of 4.9 days off per year because of illness. The manager of a large chain of grocery stores wants to know if the employees at the grocery store, on average, take fewer days off than the national average. The manager selects a random sample of 80 employees in the company and found the sample mean number of days off for the 80 employees was 4.75 days with a standard deviation of 0.9 days. When the manager performed a significance test, the P-value was 0.07. What is the meaning of this P-value in context?Dean Halverson recently read that full-time college students study 20 hours each week. She decides to do a study at her university to see if there is evidence that students study an average of less than 20 hours each week. A random sample of 30 students were asked to keep a diary of their activities over a period of several weeks. It was found that he average number of hours that the 30 students studied each week was 18.7 hours. The sample standard deviation of 3.4 hours. Find the p-value. The p-value should be rounded to 4-decimal places.The mean finish time for a yearly amateur auto race was 186.84 minutes with a standard deviation of 0.327 minute. The winning car, driven by Dan, finished in 186.48 minutes. The previous year's race had a mean finishing time of 112.4 with a standard deviation of 0.125 minute. The winning car that year, driven by Beth, finished in 112.14 minutes. Find their respective z-scores. Who had the more convincing victory? Dan had a finish time with a z-score of Beth had a finish time with a z-score of (Round to two decimal places as needed.)
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 44 participants in the treatment group lowered their cholesterol levels by a mean of 18.7 points with a standard deviation of 3.3 points. The 39 participants in the control group lowered their cholesterol levels by a mean of 18.1 points with a standard deviation of 2.1 points. Assume that the population variances are not equal and test the company’s claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha : μ1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯μ2 Step 2 of 3: what is the test statistic Step 3 of 3: draw a conclusion, fail or reject. Is…A study was being conducted to see if people who have no college degree have a lower mean salary than those who do have a college degree. A random sample of 70 people with no college degree was taken and their mean salary was $36,950 with a standard deviation of $4,375. A random sample of 75 people with a college degree was taken and their mean salary was $41,520 with a standard deviation of $3,415. Determine the appropriate formula and calculator function that would be used to test the claim that people with no college degree have a lower mean salary at the α = 0.05 level of significance. d T-Test sa/√n Fo Zo to = 2- SampFTest P1-P₂ 1 1 √p(1-6), + n₁ n₂ (₁-x₂)-(H₁-H₂) 51 $2 + n₁ n₂ 2-PropZTest 2-SampTTestA local church parish wants to raise money to add to their campus. In a sample from a previous fund raising campaign, the parish found that of the 131 people in the sample they contacted, that 69 actually contributed money. Of those that contributed money, the average contribution was $857 with a standard deviation of $265. In the recent fundraising campaign, a sample of 95 people revealed that 48 have contributed money with an average contribution of $1109 and a standard deviation of $339. When testing (at the 5% level of significance) whether the average amount contributed has increased in the current campaign compared to the previous campaign, what is the test statistic? (please round your answer to 2 decimal places)
- A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 46 participants in the treatment group lowered their cholesterol levels by a mean of 19.5 points with a standard deviation of 4.9 points. The 37 participants in the control group lowered their cholesterol levels by a mean of 18.4 points with a standard deviation of 4.1 points. Assume that the population variances are not equal and test the company's claim at the 0.02 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.A newly hired basketball coach promised a high- paced attack that will put more points on the board than the team's previously tepid offense historically managed. After a few months, the team owner looks at the data to test the coach's claim. He takes a sample of 36 of the team's games under the new coach and finds that they scored an average of 101 points with a standard deviation of 6 points. Over the past 10 years, the team had averaged 99 points. What is(are) the appropriate critical value(s) to test the new coach's claim at the 1% significance level? -2.438 -2.438 and 2.438 2.326 2.438The owners of the Metro Cash & Carry wished to study customers’ shopping habits. From earlier studies, the owners were under the impression that a typical shopper spends 0.7 hour at the mall, with a standard deviation of 0.10 hour. Recently the owners added some specialty restaurants designed to keep shoppers in the mall longer. A research team was hired to evaluate the effects of the restaurants. A sample of 45 shoppers by the team revealed that the mean time spent in the mall had increased to 0.80 hour. i. Develop a hypothesis test to determine if the mean time spent in the mall changed. Use the .10 significance level. ii. Suppose the mean shopping time actually increased from 0.75 hour to 0.79 hours. What is the probability of making a Type II error? iii)When research team reported the information in part (b) to the mall owners, the owners believed that the probability of making a Type II error was too high. How could this probability be reduced? plz sove it…
- In a certain city, the average 20- to 29-year old man is 72.5 inches tall, with a standard deviation of 3.2 inches, while the average 20- to 29-year old woman is 64.3 inches tall, with a standard deviation of 3.9 inches. Who is relatively taller, a 75-inch man or a 70-inch woman?The administrator at your local hospital states that on weekends the average wait time for emergency room visits is 10 minutes. Based on discussions you have had with friends who have complained on how long they waited to be seen in the ER over a weekend, you dispute the administrator’s claim. You decide to test your hypothesis. Over the course of a few weekends, you record the wait time for 40 randomly selected patients. The average wait time for these 40 patients is 11 minutes with a standard deviation of 3 minutes. Do you have enough evidence to support your hypothesis that the average ER wait time is different than 10 minutes? You opt to conduct the test at an α = 0.025 level of significance.A pharmaceutical company needs to know if its new cholesterol drug, Praxor, is effective at lowering cholesterol levels. It believes that people who take Praxor will average a greater decrease in cholesterol level than people taking a placebo. After the experiment is complete, the researchers find that the 34 participants in the treatment group lowered their cholesterol levels by a mean of 22.2 points with a standard deviation of 3.4 points. The 42 participants in the control group lowered their cholesterol levels by a mean of 21.2 points with a standard deviation of 1.8 points. Assume that the population variances are not equal and test the company's claim at the 0.10 level. Let the treatment group be Population 1 and let the control group be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho:₁ = ₂ Ha: M •M₂
